ABCD is a trapezium such that AB and CD are parallel and BC \bot CD. If \angleADB = θ, BC = p and CD = q, then AB is equal to :

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

ABCD is a trapezium such that AB and CD are parallel and BC $⊥$ CD. If $∠$ ADB = θ, BC = p and CD = q, then AB is equal to :

$\frac{(p^{2} + q^{2}) sinθ}{{(pcosθ + qsinθ)}^{2}}$

$\frac{p^{2} + q^{2} cosθ}{pcosθ + qsinθ}$

$\frac{p^{2} + q^{2}}{p^{2} cosθ + q^{2} sinθ}$

$\frac{(p^{2} + q^{2}) sinθ}{pcosθ + qsinθ}$

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From quadrilateral ABCD

$\frac{π}{2} + \frac{π}{2} + θ + α + A = 2 π$

$∴ A = (π - θ - α)$

$∴$ $sinA = \sin (θ + α) = sinθcosα + cosθsinα$

sin A = sin $θ$

$(\frac{q}{\sqrt{p^{2} + q^{2}}}) + cosθ (\frac{p}{\sqrt{p^{2} + q^{2}}}) (from ΔBCD)$

Apply Sine rule in $∆$ ABD

$\frac{AB}{sinθ} = \frac{\sqrt{p^{2} + q^{2}}}{sinA}$

$AB = \frac{\sqrt{p^{2} + q^{2}} \cdot (sinθ)}{sinA} = \frac{(p^{2} + q^{2}) sinθ}{pcosθ + qsinθ}$

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